A car starts from rest and moves with uniform acceleration 'a' on a straight road from time t=0 to t=T. After that, a constant deceleration brings it to rest. In this process the average speed of the car is - (a) aT/4 (b) 3aT/2 (c) aT/2 (d) aT
Answers
Answer:
aT/2
Explanation:
Hope this will help you
Answer:
From the given options, the correct option is C) a T/2
Explanation:
V = a T/2
Explanation: We are aware that
total distance travelled / total time taken = average speed
The car begins at rest and then accelerates with an acceleration 'a' after a time 'T' to a maximum velocity 'v'. Then it decelerates with the same acceleration and comes to a complete stop in time 'T' more. Let the distance travelled by it in the first case be s and the distance travelled in the second case be s'.
so, average speed will be = (s + s') / (T+T)
We already know that v2 - u2 = 2as, so
s = v2/2a
similarly
s' = v2/2a
We get from v = u + a (for the first case)
v = aT (t = T)
As a result of using this value of v in the preceding two equations, we obtain
2a2T2/a = 2aT2 s = s'
the average velocity is now
V = (s + s')/2T = (aT2 + aT2)/2T = (aT2 + aT2) / 2T
as a result,
V = aT/2
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